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In a cam-follower, the follower rises by h as the cam rotates by d (radians) at constant angular velocity w (radians/s). The follower is uniformly accelerating during the first half of the rise period and it is uniformly decelerating in the later half of the rise period. Assuming that the magnitudes of the acceleration and deceleration are same, the maximum velocity of the follower is
[2018]
- a)4hω/δ
- b)hω
- c)2ωh/δ
- d)2hω
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
In a cam-follower, the follower rises by h as the cam rotates by d (ra...
Here, out stroke angle
θo = δ
and stroke length = h
Angular velocity = ω
V = u + at
...(1)
...(2)
From (1) & (2), we get
θo = δ
and stroke length = h
Angular velocity = ω
V = u + at
...(1) ...(2)From (1) & (2), we get
Most Upvoted Answer
In a cam-follower, the follower rises by h as the cam rotates by d (ra...
B) 2h
c) h
d) h/2
The correct answer is b) 2h.
To solve this problem, we can use the equations of motion for uniformly accelerated motion. Let's denote the time taken for the follower to rise by h as t.
During the first half of the rise period, the follower is uniformly accelerating. The distance covered during this time is h/2, and the time taken is t/2. Using the equation s = ut + (1/2)at^2, where s is the distance covered, u is the initial velocity, a is the acceleration, and t is the time taken, we can write:
h/2 = 0 + (1/2)a(t/2)^2
h/2 = (1/8)at^2
Similarly, during the second half of the rise period, the follower is uniformly decelerating. The distance covered during this time is also h/2, and the time taken is t/2. Using the same equation, we can write:
h/2 = 0 + (1/2)(-a)(t/2)^2
h/2 = (1/8)(-a)t^2
Now, we can solve these two equations simultaneously to find the value of t. Dividing the second equation by the first equation, we get:
1 = -t^2/t^2
1 = -1
This is not possible, so we made an error in our calculations. Let's try again.
During the first half of the rise period, the distance covered is h/2, and the time taken is t/2. Using the equation s = ut + (1/2)at^2, we can write:
h/2 = 0 + (1/2)a(t/2)^2
h/2 = (1/8)at^2
Similarly, during the second half of the rise period, the distance covered is also h/2, but the time taken is t/2. Using the same equation, we can write:
h/2 = 0 + (1/2)(-a)(t/2)^2
h/2 = (1/8)(-a)t^2
Now, we can solve these two equations simultaneously to find the value of t. Dividing the second equation by the first equation, we get:
1 = -t^2/t^2
1 = -1
Again, this is not possible. It seems that we have made an error in our calculations.
Let's try a different approach. Since the magnitudes of the acceleration and deceleration are the same, we can assume that the follower takes equal time to accelerate and decelerate. Therefore, the time taken for the follower to rise by h is 2t.
During the first half of the rise period, the follower is uniformly accelerating. The distance covered during this time is h/2, and the time taken is t. Using the equation s = ut + (1/2)at^2, we can write:
h/2 = 0 + (1/2)at^2
Simplifying this equation, we have:
h = at^2/2
Now, we can solve this equation for a:
a = (2h)/t^2
The maximum velocity of the follower occurs
c) h
d) h/2
The correct answer is b) 2h.
To solve this problem, we can use the equations of motion for uniformly accelerated motion. Let's denote the time taken for the follower to rise by h as t.
During the first half of the rise period, the follower is uniformly accelerating. The distance covered during this time is h/2, and the time taken is t/2. Using the equation s = ut + (1/2)at^2, where s is the distance covered, u is the initial velocity, a is the acceleration, and t is the time taken, we can write:
h/2 = 0 + (1/2)a(t/2)^2
h/2 = (1/8)at^2
Similarly, during the second half of the rise period, the follower is uniformly decelerating. The distance covered during this time is also h/2, and the time taken is t/2. Using the same equation, we can write:
h/2 = 0 + (1/2)(-a)(t/2)^2
h/2 = (1/8)(-a)t^2
Now, we can solve these two equations simultaneously to find the value of t. Dividing the second equation by the first equation, we get:
1 = -t^2/t^2
1 = -1
This is not possible, so we made an error in our calculations. Let's try again.
During the first half of the rise period, the distance covered is h/2, and the time taken is t/2. Using the equation s = ut + (1/2)at^2, we can write:
h/2 = 0 + (1/2)a(t/2)^2
h/2 = (1/8)at^2
Similarly, during the second half of the rise period, the distance covered is also h/2, but the time taken is t/2. Using the same equation, we can write:
h/2 = 0 + (1/2)(-a)(t/2)^2
h/2 = (1/8)(-a)t^2
Now, we can solve these two equations simultaneously to find the value of t. Dividing the second equation by the first equation, we get:
1 = -t^2/t^2
1 = -1
Again, this is not possible. It seems that we have made an error in our calculations.
Let's try a different approach. Since the magnitudes of the acceleration and deceleration are the same, we can assume that the follower takes equal time to accelerate and decelerate. Therefore, the time taken for the follower to rise by h is 2t.
During the first half of the rise period, the follower is uniformly accelerating. The distance covered during this time is h/2, and the time taken is t. Using the equation s = ut + (1/2)at^2, we can write:
h/2 = 0 + (1/2)at^2
Simplifying this equation, we have:
h = at^2/2
Now, we can solve this equation for a:
a = (2h)/t^2
The maximum velocity of the follower occurs
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In a cam-follower, the follower rises by h as the cam rotates by d (radians) at constant angular velocity w (radians/s). The follower is uniformly accelerating during the first half of the rise period and it is uniformly decelerating in the later half of the rise period. Assuming that the magnitudes of the acceleration and deceleration are same, the maximum velocity of the follower is[2018]a)4hω/δb)hωc)2ωh/δd)2hωCorrect answer is option 'C'. Can you explain this answer?
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In a cam-follower, the follower rises by h as the cam rotates by d (radians) at constant angular velocity w (radians/s). The follower is uniformly accelerating during the first half of the rise period and it is uniformly decelerating in the later half of the rise period. Assuming that the magnitudes of the acceleration and deceleration are same, the maximum velocity of the follower is[2018]a)4hω/δb)hωc)2ωh/δd)2hωCorrect answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about In a cam-follower, the follower rises by h as the cam rotates by d (radians) at constant angular velocity w (radians/s). The follower is uniformly accelerating during the first half of the rise period and it is uniformly decelerating in the later half of the rise period. Assuming that the magnitudes of the acceleration and deceleration are same, the maximum velocity of the follower is[2018]a)4hω/δb)hωc)2ωh/δd)2hωCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a cam-follower, the follower rises by h as the cam rotates by d (radians) at constant angular velocity w (radians/s). The follower is uniformly accelerating during the first half of the rise period and it is uniformly decelerating in the later half of the rise period. Assuming that the magnitudes of the acceleration and deceleration are same, the maximum velocity of the follower is[2018]a)4hω/δb)hωc)2ωh/δd)2hωCorrect answer is option 'C'. Can you explain this answer?.
In a cam-follower, the follower rises by h as the cam rotates by d (radians) at constant angular velocity w (radians/s). The follower is uniformly accelerating during the first half of the rise period and it is uniformly decelerating in the later half of the rise period. Assuming that the magnitudes of the acceleration and deceleration are same, the maximum velocity of the follower is[2018]a)4hω/δb)hωc)2ωh/δd)2hωCorrect answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about In a cam-follower, the follower rises by h as the cam rotates by d (radians) at constant angular velocity w (radians/s). The follower is uniformly accelerating during the first half of the rise period and it is uniformly decelerating in the later half of the rise period. Assuming that the magnitudes of the acceleration and deceleration are same, the maximum velocity of the follower is[2018]a)4hω/δb)hωc)2ωh/δd)2hωCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a cam-follower, the follower rises by h as the cam rotates by d (radians) at constant angular velocity w (radians/s). The follower is uniformly accelerating during the first half of the rise period and it is uniformly decelerating in the later half of the rise period. Assuming that the magnitudes of the acceleration and deceleration are same, the maximum velocity of the follower is[2018]a)4hω/δb)hωc)2ωh/δd)2hωCorrect answer is option 'C'. Can you explain this answer?.
Solutions for In a cam-follower, the follower rises by h as the cam rotates by d (radians) at constant angular velocity w (radians/s). The follower is uniformly accelerating during the first half of the rise period and it is uniformly decelerating in the later half of the rise period. Assuming that the magnitudes of the acceleration and deceleration are same, the maximum velocity of the follower is[2018]a)4hω/δb)hωc)2ωh/δd)2hωCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering.
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